Coal miners often find mice in deep mines but rarely find rats; let’s see if we can figure out why. A mouse is roughly 5 cm long by 2 cm wide and has a mass of 30 g; a rat is roughly 20 cm long by 5 cm wide and has a mass of 500 g. Assume that both have a drag coefficient CD ≈ 0.3.
- a) Estimate the terminal falling speeds reached by a mouse and a rat, respectively.
- b) Assume that mine shafts are deep enough that both the mouse and the rat reach terminal velocity before hitting the bottom. Estimate the magnitude of the maximum force required to stop both the rat and the mouse when they hit the bottom. (Hint: Over what distance will the center of mass travel between the beginning of the collision and when the animal is at rest? What is the acceleration required to bring the animal to rest over this distance? The center of mass can’t travel more than about half the height of an animal, why?)
- c) Bones will break if they are subjected to a compressional force per unit area of more than about 1.5 × 108 N/m2. A mouse may have leg bones about 1.5 mm in diameter; for a rat, they might be about twice as thick. Using your estimates from part b), determine if either the mouse or the rat (or both) will suffer broken legs. Is your answer consistent with the observations of the coal miners? Explain.
FD = ρ*v^2*CD*A/2
F = m*a
xcm = (m1*r1 + m2*r2 + ... + mn*rn)/(m1 + m2 + ... + mn)
The Attempt at a Solution
I did part A without a problem. Simply plugged in values into the FD equation and set it to equal mg for each: mouse and rat. Mouse terminal velocity: 44.27 m/s, Rat terminal velocity: 57.155 m/s.
Part B, I'm not too sure about it. I have no idea what the hints really mean. I actually have no clue how to even start it.
Part C, using Part B's answer, I feel like I can figure it out, however, I'm not entirely sure what area to use, as it states that the leg bone is 1.5 mm in diameter, but says nothing about length, so I'm not sure what to use for surface area.